Mobius band is differentiable manifold

Question

Let $X:=[0,1]\times\mathbb{R}$ be equipped with the subspace topology of $\mathbb{R}^2$ and let $(0,t)\sim(1,-t)$ be an equivalence relation on $X$. I must show that the quotient $M:=X/\sim$ equipped with the quotient topology is a differentiable manifold.

I understand that each equivalence class consists of two points:

$[(x,y)]=\{(x,y);(1-x,-y)\}$, i.e. each pair of points corresponding to a reflection at $(\frac{1}{2},0)$.

I first want to show that $M$ is indeed a topological manifold. It is non-empty and a topological space, as is evident. I want to prove that $M$ is hausdorff by showing that $M$ is homeomorphic to $M_{\geq0}:=[0,1]\times\mathbb{R}_{\geq0}$. The space $M_{\geq0}$ is itself hausdorff as a subspace of $\mathbb{R}^2$, if we equip it with the subspace topology.

We can construct the following map.

$\rho:M\rightarrow M_{\geq0}; [(x,y)]\mapsto\begin{cases}(x,y),\quad y\geq0\\ (1-x,-y),\quad y<0\end{cases}$

It is injective:

$\rho([(x,y)])=\rho([(u,v)])\Rightarrow (x,y)=(u,v)\vee(1-x,-y)=(1-u,-v)$,

from which both follows that $(x,y)=(u,v)$.

It is surjective:

Let $x\in[0,1]$ and $y\in\mathbb{R}$.

In the case $y\geq 0$, we obtain $\rho([(x,y)])=(x,y)\in[0,1]\times\mathbb{R}_{\geq0}=M_{\geq0}$.

In the case $y<0$, we obtain $\rho([(x,y)])=(1-x,-y)$,

where $1-x\in[0,1]$ and $-y\in\mathbb{R}_{\geq0}$, hence $\rho([(x,y)])\in[0,1]\times\mathbb{R}_{\geq0}=M_{\geq0}$.

Thus, in all we have $\rho(M)=M_{\geq0}$ and we have shown that $\rho$ is bijective.

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Now, how do I show that $\rho$ is continuous? I tried to think of what the open sets from the quotient topology of $M$ look like but that seems to be too complicated an approach because ther are so many cases to be considered because of the subspace topology on $X$ and the corresponding different forms of open sets in $M$. Where is the error in my thoughts?

Answer 1

The idea is to use separating neighbourhoods for representatives of classes $[x] \neq [y]$ in $M$. We can split up the proof in three case: a) $x,y$ both lie in the interior of $X$; b) One of $x$,$y$ lies in the interior of $X$ and the other lies on the boundary; c) Both $x$ and $y$ lie on the boundary of $X$.

a) In this case, pick separating neighbourhoods $U,V$ for $x,y$ respectively that are small enough that they don't touch the boundary. Then since the equivalence relation leaves the interior of $X$ untouched, $\pi(U) \cap \pi(V) = \pi(U \cap V) = \emptyset$.

b) Without loss of generality, we may assume that $x \in \partial X$ and $y \not\in \partial X$. We may find an $\epsilon > 0$ small enough so that $y$ is not contained in the fattened boundary $U = [0,\epsilon]\times \mathbb R \cup [1-\epsilon,1] \times \mathbb R$. Since the complement of $U$ is open in $X$, we may then find a neighborhood $V$ of $y$ that also doesn't intersect $U$. The equivalence relation will leave $V$ intact and perturb the part of $U$ that lies on the boundary, and overall we will have $$\begin{align}\pi(U) \cap \pi(V) &= (\pi(\partial X) \cup \pi(U \setminus \partial X)) \cap \pi(V) \\&= (\pi(\partial X) \cap \pi(V)) \cup (\pi(U \setminus \partial X) \cap \pi(V))\\ &= \emptyset \cup \pi((U \setminus \partial X) \cap V)\\&= \emptyset.\end{align} $$

c) If $x,y \in \partial X$, we have to make sure that the separating neighbourhoods in $X$ are small enough so that they don't get squished together in $M$. For example, a tall and skinny open neighbourhood of $(0,1)$ that contains $(0,-1)$ will, down in $M$, intersect a tall and skinny open neighbourhood of $(1,1)$ that contains $(1,-1)$, even though they don't intersect in $X$. The idea, then, is to make sure that the neighbourhoods are not too tall. To this end, choose the representatives $x,y$ so that their first coordinate is $0$, i.e., they both lie on the left boundary of $X$. Then any $U,V$ that separate $x,y$ and don't intersect the right boundary of $X$ will do the job, because the parts of $U,V$ that don't intersect the left boundary boundary will be unchanged by the projection, and the parts that intersect the left boundary won't get mixed up by the projection (as $\pi$ is injective when restricted to $X$ minus the right boundary). I'll leave it to you to write this down as formally as you want to.

Answer 2

Firstly the mobius strip $M$ can be defined as the quotient space $M = ([0, 1] \times [0, 1])/\sim $ generated by the relation $(x, 0) \sim (1-x, 1)$ for $0 \leq x \leq 1$.

From your question I take it that you want to prove that $M$ is Hausdorff. There is a much simpler way to do this then just by brute-force proving it by the definition of what it means for a topological space to be Hausdorff. (But credits to you for attempting to prove it this way anyhow)

Theorem: The quotient space of a compact Hausdorff space is Hausdorff.

Note now that $[0,1]$ is compact and also Hausdorff, so the product $[0, 1] \times [0,1]$ is compact and also Hausdorff. Since $M$ is the quotient space of $[0, 1] \times [0, 1]$ under a specific relatation by the theorem above we have that $M$ is Hausdorff.

Answer 3

So, I wanted to show that $M$ is hausdorff. The approach with the homeomorphism did not help. After some meddling around I came to the solution.

I pick $[x],[y]\in M$ with $[x]\neq[y]$. First, I have to think of what preimages in $X$ fulfill this requirement. One finds that

$$[x]\neq[y]\Rightarrow\begin{cases}x\neq y\wedge x\neq\bar{y}\quad x,y\in\partial X\\ x\neq y,\quad else.\end{cases}$$ where $\bar{x}=\overline{(x_1,x_2)}:=(1-x_1,-x_2)$ is the reflection of $x$ at $(1/2,0)$.

Now for each case we pick such corresponding $x,y\in X$. Then we have $[x]\neq[y]$. Since, in either case we have $x\neq y$, we can find separating open neighbourhoods $U(x), V(y)$ with $U\cap V=\emptyset$. From this follows that $\pi(U\cap V)=\pi(U)\cap\pi(V)=\emptyset$. Because $M$ is equipped with the quotient topology and $U,V\subset X$ are open and $[x]\in\pi(U),[y]\in\pi(V)$, we have separating open neighbourhoods to each $[x]\neq[y]$.

Hence, $M$ is hausdorff, q.e.d..